Introduction to Probability and Statistics for Engineers and Scientists

146 Chapter 5: Special Random Variables

EXAMPLE 5.1d Suppose that 10 percent of the chips produced by a computer hardware
manufacturer are defective. If we order 100 such chips, willX, the number of defective
ones we receive, be a binomial random variable?

SOLUTION The random variableXwill be a binomial random variable with parameters
(100, .1) if each chip has probability .9 of being functional and if the functioning of
successive chips is independent. Whether this is a reasonable assumption when we know
that 10 percent of the chips produced are defective depends on additional factors. For
instance, suppose that all the chips produced on a given day are always either functional
or defective (with 90 percent of the days resulting in functional chips). In this case, if we
know that all of our 100 chips were manufactured on the same day, thenXwill not be
a binomial random variable. This is so since the independence of successive chips is not
valid. In fact, in this case, we would have

P{X= 100 }=.1

P{X= 0 }=.9 ■

Since a binomial random variableX, with parametersnandp, represents the number of
successes innindependent trials, each having success probabilityp, we can representXas
follows:

X=

∑n

i= 1

Xi (5.1.3)

where

Xi=

{

1 if theith trial is a success

0 otherwise

Because theXi,i=1,...,nare independent Bernoulli random variables, we have that

E[Xi]=P{Xi= 1 }=p

Var(Xi)=E[Xi^2 ]−p^2

=p(1−p)

where the last equality follows sinceXi^2 =Xi, and soE[Xi^2 ]=E[Xi]=p.
Using the representation Equation 5.1.3, it is now an easy matter to compute the mean
and variance ofX:

E[X]=

∑n

i= 1

E[Xi]

=np